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(2\(x\) - 1)3 = 125
(2\(x\) - 1)3 = 53
2\(x\) - 1 = 5
2\(x\) = 6
\(x\) = 3
(2\(x\) - 1)5 = \(x^5\)
2\(x\) - 1 = \(x\)
2\(x\) - \(x\) = 1
\(x\) = 1
2\(x^5\) + 2 = 4
2\(x^5\) = 4 - 2
2\(x^5\) = 2
\(x^5\) = 1
\(x\) = 1
\(a,128-3.\left(x+4\right)=23\\ \Rightarrow3.\left(x+4\right)=105\\ \Rightarrow x+4=35\\ \Rightarrow x=31\\ b,\left[\left(4x+28\right).3+55\right]:5=35\\ \Rightarrow\left(4x+28\right).3+55=175\\ \Rightarrow4x+28.3=120\\ \Rightarrow4x+28=60\\ \Rightarrow4x=32\\ \Rightarrow x=8.\)
c) \(\left(12x-4^3\right).8^3=4.8^4\)
\(12x-64=4.8^4:8^3\)
\(12x-64=32\)
\(12x=32+64\)
\(12x=96\)
\(x=\dfrac{96}{12}\)
\(x=8\)
d) \(720:\left[41-\left(2x-5\right)\right]:5=35\)
\(720:\left(41-2x+5\right):5=35\)
\(720:\left(46-2x\right)=35.5\)
\(720:\left(46-2x\right)=175\)
\(46-2x=720:175\)
\(46-2x=\dfrac{144}{35}\)
\(2x=46-\dfrac{144}{35}\)
\(2x=\dfrac{1466}{35}\)
\(x=\dfrac{1466}{35}:2\)
\(x=\dfrac{733}{35}\)
a, 128 – 3(x+4) = 23
b, 12 x - 4 3 . 8 3 = 4 . 8 4
c, [(4x+28).3+55]:5 = 35
d, 720:[41 – (2x – 5)] = 2 3 . 5
\(16^x< 128^4\)
=> \(\left[2^4\right]^x< \left[2^7\right]^4\)
=> \(2^{4x}< 2^{28}\)
=> 4x < 28
=> x < 7
Đến đây tìm x được rồi
\(\left[3x^2-5\right]+3^4+6^0=5^3\)
=> \(\left[3x^2-5\right]=5^3-6^0-3^4=43\)
=> \(3x^2-5=43\)
=> \(3x^2=48\)
=> \(x^2=16\)
=> \(x=\pm4\)
\(3x+2x\left[2^3\cdot5-3^2\cdot4\right]+5^2=4^4\)
=> \(3x+2x\left[8\cdot5-9\cdot4\right]+25=256\)
=> \(3x+2x\cdot4+25=256\)
=> \(3x+2x\cdot4=231\)
Đến đây tìm x
128-3.(x+4) =23
3.(x+4) =128-23
3.(x+4) = 105
x+4 =105:3
x+4 =35
x = 35-4
x=31
a,
128-3x-12=23
3x=128-12-23
3x=93
x=93:3
= 31
b,
(12x+84+55):5=35
12x+84+55=35.5
12x+84+55=175
12x=175-55-84
12x=36
x=36:12
x=3
<=>128-3x-12=23
<=>-3x=23+12-128
<=>-3x=-93
<=>x=31
`@` `\text {Ans}`
`\downarrow`
`a)`
`128-3*(x+4)=23`
`=> 3*(x + 4) = 128 - 23`
`=> 3*(x + 4) = 105`
`=> x + 4 = 105 \div 3`
`=> x + 4 = 35`
`=> x = 35 - 4`
`=> x = 31`
Vậy, `x = 31.`